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(6r+5)(2r+7)=0
We multiply parentheses ..
(+12r^2+42r+10r+35)=0
We get rid of parentheses
12r^2+42r+10r+35=0
We add all the numbers together, and all the variables
12r^2+52r+35=0
a = 12; b = 52; c = +35;
Δ = b2-4ac
Δ = 522-4·12·35
Δ = 1024
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1024}=32$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(52)-32}{2*12}=\frac{-84}{24} =-3+1/2 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(52)+32}{2*12}=\frac{-20}{24} =-5/6 $
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